If we had a case where we are looping over the array of numbers and the numbers can scale to infinity, and inside of each iteration we are looping over digits of each number, so for number 125556 we would loop through six numbers, 1, 2, 5, 5, 5, and 6, is the time complexity of this algorithm just O(N), where N represents the numbers in the array, or it is O(N*K) where K is the number of digits in the number. In this case K is always less than N, so I am unclear whether this is multiplication or we can just disregard the number of digits?
The algorithm you describe is always O(N * K).
However, if you know something about the relationship between N and K, then you can simplify the expression. For instance, if the numbers are guaranteed to fit in a 32-bit integer representation on a computer, then K is a constant and your algorithm is O(N). If K < N, then you can say O(N2).
But if you have no assumption on K, then you have to go with O(N * K), as K could be significantly larger than N or vice versa. Intuitively, your time complexity depends on two factors, so you need to express it with two variables unless they depend on each other.
Edit:
Since you clarified that you are looping through the numbers in order, as in 1, 2, ..., N, we now have some information on the relationship between K and N. In fact, K = O(logN), so the algorithm can be expressed as O(N logN).
If you are confused about how we know that K = O(logN), then take any power of 10 as an example. You will find that 10K has log10 10K + 1 = K + 1 digits. Similarly, any number X has O(log X) decimal digits (notice that the base of the logarithm does not matter in the big-O notation).
Related
So I have two base 10 int number with m digits and n digits. What should be the time complexity of m x n? I have done some lookups and saw that in background, the two numbers are converted into binary numbers and you add the sum of all product between each digit of one number with the other number. So will it be O(m x n) or O( numDigitsInBinaryOfNumberOne * numDigitsInBinaryOfNumberTwo) ?
This question already has answers here:
Why is the constant always dropped from big O analysis?
(7 answers)
Closed 2 years ago.
I commonly use a site called LeetCode for practice on problems. On a lot of answers in the discuss section of a problem, I noticed that run times like O(N + N) or O(2N) gets changed to O(N). For example:
int[] nums = {1, 2, 3, 4, 5};
for(int i = 0; i < nums.length; i ++) {
System.out.println(nums[i]);
}
for(int i = 0; i < nums.length; i ++) {
System.out.println(nums[i]);
}
This becomes O(N), even though it iterates through nums twice. Why is it not O(2N) or O(N + N)?
In time complexity, constant coefficients do not play a role. This is because the actual time it takes an algorithm to run depends also on the physical constraints of the machine. This means that if you run your code on a machine which is twice as fast as another, all other conditions being equal, it would run in about half the time with the same input.
But that’s not the same thing when you compare two algorithms with different time complexities. For example, when you compare the running time of an algorithm of O( N ^ 2 ) to an algorithm of O(N), the running time of O( N ^ 2 ) grows so fast with the growth of input size that the O(N) one cannot catch up with it, no matter how big you choose its constant coefficient.
Let’s say your constant coefficient is 1000, instead of just 2, then for input sizes of ( N > 1000 ) the running time of O( N ^ 2 ) algorithm becomes a proportion of ( N * N ) while N would be growing proportional to the input size, while the running time of the O(N) algorithm only remains proportional to ( 1000 * N ).
Time complexity for O(n+n) reduces to O(2n). Now 2 is a constant. So the time complexity will essentially depend on n.
Hence the time complexity of O(2n) equates to O(n).
Also if there is something like this O(2n + 3) it will still be O(n) as essentially the time will depend on the size of n.
Now suppose there is a code which is O(n^2 + n), it will be O(n^2) as when the value of n increases the effect of n will become less significant compared to effect of n^2.
I have an integer, N.
I denote f[i] = number of appearances of the digit i in N.
Now, I have the following algorithm.
FOR i = 0 TO 9
FOR j = 1 TO f[i]
k = k*10 + i;
My teacher said this is O(N). It seems to me more like a O(logN) algorithm.
Am I missing something?
I think that you and your teacher are saying the same thing but it gets confused because the integer you are using is named N but it is also common to refer to an algorithm that is linear in the size of its input as O(N). N is getting overloaded as the specific name and the generic figure of speech.
Suppose we say instead that your number is Z and its digits are counted in the array d and then their frequencies are in f. For example, we could have:
Z = 12321
d = [1,2,3,2,1]
f = [0,2,2,1,0,0,0,0,0,0]
Then the cost of going through all the digits in d and computing the count for each will be O( size(d) ) = O( log (Z) ). This is basically what your second loop is doing in reverse, it's executing one time for each occurence of each digits. So you are right that there is something logarithmic going on here -- the number of digits of Z is logarithmic in the size of Z. But your teacher is also right that there is something linear going on here -- counting those digits is linear in the number of digits.
The time complexity of an algorithm is generally measured as a function of the input size. Your algorithm doesn't take N as an input; the input seems to be the array f. There is another variable named k which your code doesn't declare, but I assume that's an oversight and you meant to initialise e.g. k = 0 before the first loop, so that k is not an input to the algorithm.
The outer loop runs 10 times, and the inner loop runs f[i] times for each i. Therefore the total number of iterations of the inner loop equals the sum of the numbers in the array f. So the complexity could be written as O(sum(f)) or O(Σf) where Σ is the mathematical symbol for summation.
Since you defined that N is an integer which f counts the digits of, it is in fact possible to prove that O(Σf) is the same thing as O(log N), so long as N must be a positive integer. This is because Σf equals how many digits the number N has, which is approximately (log N) / (log 10). So by your definition of N, you are correct.
My guess is that your teacher disagrees with you because they think N means something else. If your teacher defines N = Σf then the complexity would be O(N). Or perhaps your teacher made a genuine mistake; that is not impossible. But the first thing to do is make sure you agree on the meaning of N.
I find your explanation a bit confusing, but lets assume N = 9075936782959 is an integer. Then O(N) doesn't really make sense. O(length of N) makes more sense. I'll use n for the length of N.
Then f(i) = iterate over each number in N and sum to find how many times i is in N, that makes O(f(i)) = n (it's linear). I'm assuming f(i) is a function, not an array.
Your algorithm loops at most:
10 times (first loop)
0 to n times, but the total is n (the sum of f(i) for all digits must be n)
It's tempting to say that algorithm is then O(algo) = 10 + n*f(i) = n^2 (removing the constant), but f(i) is only calculated 10 times, each time the second loops is entered, so O(algo) = 10 + n + 10*f(i) = 10 + 11n = n. If f(i) is an array, it's constant time.
I'm sure I didn't see the problem the same way as you. I'm still a little confused about the definition in your question. How did you come up with log(n)?
I have some difficulties in understanding the time complexity analysis for one solution for the Happy Number Question from Leet code, for my doubts on complexity analysis, I marked them in bold and really appreciate your advice
Here is the question:
Link: https://leetcode.com/problems/happy-number/
Question:
Write an algorithm to determine if a number is "happy".
A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.
Example:
Input: 19
Output: true
Explanation:
1^2(square of 1) + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1
Here is the code:
class Solution(object):
def isHappy(self, n):
#getnext function will compute the sum of square of each digit of n
def getnext(n):
totalsum = 0
while n>0:
n,v = divmod(n,10)
totalsum+=v**2
return totalsum
#we declare seen as a set to track the number we already visited
seen = set()
#we stop checking if: either the number reaches one or the number was visited #already(ex.a cycle)
while n!=1 and (n not in seen):
seen.add(n)
n = getnext(n)
return n==1
Note: feel free to let me know if I need to explain how the code works
Time Complexity Analysis:
Time complexity : O(243 * 3 + logN + loglogN + log loglog N)...=O(logN).
Finding the next value for a given number has a cost of O(log n)because we are processing each digit in the number, and the number of digits in a number is given by logN.
My doubt: why the number of digits in a number is given by logN? what is N here? the value of a specific number or something else?
To work out the total time complexity, we'll need to think carefully about how many numbers are in the chain, and how big they are.
We determined above that once a number is below 243, it is impossible for it to go back up above 243.Therefore, based on our very shallow analysis we know for sure that once a number is below 243, it is impossible for it to take more than another 243 steps to terminate.
Each of these numbers has at most 3 digits. With a little more analysis, we could replace the 243 with the length of the longest number chain below 243, however because the constant doesn't matter anyway, we won't worry about it.
My doubt: I think the above paragraph is related to the time complexity component of 243*3, but I cannot understand why we multiply 243 by 3
For an n above 243, we need to consider the cost of each number in the chain that is above 243. With a little math, we can show that in the worst case, these costs will be O(log n) + O(log log n) + O(log log log N)... Luckily for us, the O(logN) is the dominating part, and the others are all tiny in comparison (collectively, they add up to less than logN), so we can ignore them. My doubt: what is the reasoning behind O(log log n) O(log log log N) for an n above 243?
Well, my guess for the first doubt is that the number of digits of a base 10 number is given by it's value (N) taken to the logarithm at base 10, rounded down. So for example, 1023 would have floor(log10(1023)) digits, which is 3. So yes, the N is the value of the number. the log in time complexity indicates a logarithm, not specifically that of base 2 or base e.
As for the second doubt, it probably has to do with the work required to reduce a number to below 243, but I am not sure. I'll edit this answer once I work that bit out.
Let's say N has M digits. Than getnext(N) <= 81*M. The equality happens when N only has 9's.
When N < 1000, i.e. at most 3 digits, getnext(N) <= 3*81 = 243. Now, you will have to call getnext(.) at most O(243) times to figure out if N is indeed happy.
If M > 3, number of digits of getnext(N) must be less than M. Try getnext(9999), getnext(99999), and so on [1].
Notes:
[1] Adding a digit to N can make it at most 10*N + 9, i.e. adding a 9 at the end. But the number of digits increases to M+1 only. It's a logarithmic relationship between N and M. Hence, the same relationship holds between N and 81*M.
Using the Leetcode solution
class Solution {
private int getNext(int n) {
int totalSum = 0;
while (n > 0) {
int d = n % 10;
n = n / 10;
totalSum += d * d;
}
return totalSum;
}
public boolean isHappy(int n) {
Set<Integer> seen = new HashSet<>();
while (n != 1 && !seen.contains(n)) {
seen.add(n);
n = getNext(n);
}
return n == 1;
}
}
}
O(243*3) for n < 243
3 is the max number of digits in n
e.g. For n = 243
getNext() will take a maximum of 3 iterations because there are 3 digits for us to loop over.
isHappy() can take a maximum of 243 iterations to find a cycle or terminate, because we can store a max of 243 numbers in our hash set.
O(log n) + O(log log n) + O(log log log N)... for n > 243
1st iteration + 2nd iteration + 3rd iteration ...
getNext() will be called a maximum of O(log n) times. Because log10 n is the number of digits.
isHappy() will be called a maximum of 9^2 per digit. This is the max we can store in the hash set before we find a cycle or terminate.
First Iteration
9^2 * number of digits
O(81*(log n)) drop the constant
O(log n)
+
Second Iteration
O(log (81*(log n))) drop the constant
O(log log n)
+
Third Iteration
O(log log log N)
+
ect ...
Suppose we have an algorithm that is of order O(2^n). Furthermore, suppose we multiplied the input size n by 2 so now we have an input of size 2n. How is the time affected? Do we look at the problem as if the original time was 2^n and now it became 2^(2n) so the answer would be that the new time is the power of 2 of the previous time?
Big 0 is not for telling you the actual running time, just how the running time is affected by the size of input. If you double the size of input the complexity is still O(2^n), n is just bigger.
number of elements(n) units of work
1 1
2 4
3 8
4 16
5 32
... ...
10 1024
20 1048576
There's a misunderstanding here about how Big-O relates to execution time.
Consider the following formulas which define execution time:
f1(n) = 2^n + 5000n^2 + 12300
f2(n) = (500 * 2^n) + 6
f3(n) = 500n^2 + 25000n + 456000
f4(n) = 400000000
Each of these functions are O(2^n); that is, they can each be shown to be less than M * 2^n for an arbitrary M and starting n0 value. But obviously, the change in execution time you notice for doubling the size from n1 to 2 * n1 will vary wildly between them (not at all in the case of f4(n)). You cannot use Big-O analysis to determine effects on execution time. It only defines an upper boundary on the execution time (which is not even guaranteed to be the minimum form of the upper bound).
Some related academia below:
There are three notable bounding functions in this category:
O(f(n)): Big-O - This defines a upper-bound.
Ω(f(n)): Big-Omega - This defines a lower-bound.
Θ(f(n)): Big-Theta - This defines a tight-bound.
A given time function f(n) is Θ(g(n)) only if it is also Ω(g(n)) and O(g(n)) (that is, both upper and lower bounded).
You are dealing with Big-O, which is the usual "entry point" to the discussion; we will neglect the other two entirely.
Consider the definition from Wikipedia:
Let f and g be two functions defined on some subset of the real numbers. One writes:
f(x)=O(g(x)) as x tends to infinity
if and only if there is a positive constant M such that for all sufficiently large values of x, the absolute value of f(x) is at most M multiplied by the absolute value of g(x). That is, f(x) = O(g(x)) if and only if there exists a positive real number M and a real number x0 such that
|f(x)| <= M|g(x)| for all x > x0
Going from here, assume we have f1(n) = 2^n. If we were to compare that to f2(n) = 2^(2n) = 4^n, how would f1(n) and f2(n) relate to each other in Big-O terms?
Is 2^n <= M * 4^n for some arbitrary M and n0 value? Of course! Using M = 1 and n0 = 1, it is true. Thus, 2^n is upper-bounded by O(4^n).
Is 4^n <= M * 2^n for some arbitrary M and n0 value? This is where you run into problems... for no constant value of M can you make 2^n grow faster than 4^n as n gets arbitrarily large. Thus, 4^n is not upper-bounded by O(2^n).
See comments for further explanations, but indeed, this is just an example I came up with to help you grasp Big-O concept. That is not the actual algorithmic meaning.
Suppose you have an array, arr = [1, 2, 3, 4, 5].
An example of a O(1) operation would be directly access an index, such as arr[0] or arr[2].
An example of a O(n) operation would be a loop that could iterate through all your array, such as for elem in arr:.
n would be the size of your array. If your array is twice as big as the original array, n would also be twice as big. That's how variables work.
See Big-O Cheat Sheet for complementary informations.